We have
![(-3xy^2+y)_y=--6xy+1](https://tex.z-dn.net/?f=%28-3xy%5E2%2By%29_y%3D--6xy%2B1)
and
![(-3x^2y+x)_x=-6xy+1](https://tex.z-dn.net/?f=%28-3x%5E2y%2Bx%29_x%3D-6xy%2B1)
so the equation is indeed exact. So we want to find a function
such that
![F_x=-3xy^2+y](https://tex.z-dn.net/?f=F_x%3D-3xy%5E2%2By)
![F_y=-3x^2y+x](https://tex.z-dn.net/?f=F_y%3D-3x%5E2y%2Bx)
Integrating both sides of the first equation wrt
gives
![F(x,y)=-\dfrac32x^2y^2+xy+f(y)](https://tex.z-dn.net/?f=F%28x%2Cy%29%3D-%5Cdfrac32x%5E2y%5E2%2Bxy%2Bf%28y%29)
Differentiating both sides wrt
gives
![F_y=-3x^2y+x=-3x^2y+x+f_y\implies f_y=0\implies f(y)=C](https://tex.z-dn.net/?f=F_y%3D-3x%5E2y%2Bx%3D-3x%5E2y%2Bx%2Bf_y%5Cimplies%20f_y%3D0%5Cimplies%20f%28y%29%3DC)
So we have
![F(x,y)=-\dfrac32x^2y^2+xy+C=C](https://tex.z-dn.net/?f=F%28x%2Cy%29%3D-%5Cdfrac32x%5E2y%5E2%2Bxy%2BC%3DC)
or
![F(x,y)=\boxed{-\dfrac32x^2y^2+xy=C}](https://tex.z-dn.net/?f=F%28x%2Cy%29%3D%5Cboxed%7B-%5Cdfrac32x%5E2y%5E2%2Bxy%3DC%7D)
Answer:
4568
Step-by-step explanation:
( - ∞, 3) ∪ (3, ∞ )
The domain is the set of values of x which make f(x) defined
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.
solve x - 3 = 0 ⇒ x = 3
domain : (- ∞, 3) ∪ (3, ∞ )
I believe it is the third choice.
the answer is cubic trinomial